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## Algebra (all content)

### Course: Algebra (all content)>Unit 11

Lesson 20: Advanced interpretation of exponential models (Algebra 2 level)

# Structure in exponential expression

Analyzing an elaborate exponential function to determine its value at t=0 and whether it increases or decreases with t. This is algebraic reasoning at its best! Created by Sal Khan.

## Want to join the conversation?

• . How do we know that the expression is going to be always larger than the initial position (c^2 - 1)? I understand that the whole expression will be increasing as (-d^-t) gets smaller as t increases. But how do we make sure that the expression doesn't initially decrease from the starting point and then starts to increase to its maximum? In other words, between (c^2 - 1) and (-d^-t + c^4)/(c^2 + 1), how do we prove that the latter expression is ALWAYS going to be larger than the former? • Good question! Here is why:
t is always going to be a positive number, because we have defined it as time, and we started our time at 0.
That means that -t will always be negative.
For t>0, d^-t=1/(d^t).
Since d>1, d^t will always be >1 for any t>0.
Therefore, 1/d^t will always be between 0 and 1 and -1/(d^t) will always be between -1 and 0.
So for any t>0, the formula looks like this:
(a small negative number between -1 and 0) + c^4 / (c^2+1).

Then we need to prove that that expression above is always > c^2-1.
I'm going to rename that "small negative number between -1 and 0" to something easier. Let's call it s. -1<s<0.
The formula rewritten using s becomes (s+c^4)/(c^2+1).
Because of the boundaries we set on s, c^4+s > c^4-1. Now let's divide both sides of that inequality by c^2+1.
(c^4+s)/(c^2+1)>(c^4-1)/(c^2+1).
We have already discovered that we can use difference of squares to make the right-hand side of that equation simplify to c^2-1.
So now we have (c^4+s)/(c^2+1)> c^2-1, which is what we set out to prove!
• why is this video even here?it has nothing related to reasoning about expressions.even if it does , why does the difficulty suddenly increase so much? • Indeed, I don't understand what position of particle is, any help please ? • I don´t quite get how does c^4 - 1 is assumed to be = to (c^2)^2 - 1^2 (the -1^2 part gets me confused).
Can anyone explain me that part a little better? • How are you supposed to know whether to take 'negative d raised to the t power' or 'd raise to the t power then take the negative'? Shouldn't there be parentheses like -(d^t) not -d^t? • I think I'm missing the big picture here. What are we doing during this video with the function? Thanks. • One of the viewers asked below, "What is a particle?" Well, in this video, Sal is not trying to find out the position of a car or train or plane at a certain time. If he were, the variables would be distance = rate x time or time = distance / rate (divided by rate). Instead, Sal is asking about the position of a particle at a certain time. "Particle" is a term used in physics and it refers to subatomic "particles". These are pieces of matter such as electrons, protons, neutrons, photons and quarks (the substance that makes up protons and neutrons). Think very, very, very small and moving very, very, very, very fast. So, look at the formula in the video again. We have a d-term (distance), a t-term (time) and a c-term (rate). In physics c is the speed of light (approximately 3 x 10^8 meters/second). So, we do have the same terms and the same formula that I started with above re cars, trains and planes, only our rate is in terms of the speed of light instead of mph or kmph. So, Sal is asking what is the location of a subatomic particle (take your pick , let's say a photon) that is moving somewhere close to the speed of light after so many seconds or minutes or years?

In regard to the second question that Sal asks in the video, "What happens as time increases ("t" gets larger)?", I think it is a much more straight forward explanation (easier to comprehend), if you start by simplifying the negative exponent (-t).

[To better follow this explanation, I encourage you to copy the original formula (p(t) = -d^-t +c^4 / c^2 +1 on a piece of paper and manipulate the terms as per my explanation.]

To do this just take the reciprocal of -d^-t. This has the result of putting the term -d^(-t) in the denominator as (-d^t) and (c^4 +1) in the numerator. Now, (c^4 + 1) in the numerator is a constant (doesn't change over time). And in the denominator the term (c^2 + 1) is also a constant (doesn't change over time). Since the minus d^t (we got rid of the -t) in the denominator will increase exponentially (think very quickly) over time and this increasing value is being subtracted from the constant term in the denominator (c^2 +1), the result is the denominator will be get smaller and smaller (approaching 0) as time gets larger (the passage of time). We, all know that for a fraction as the denominator gets smaller the number represented by the fraction gets larger. So, the answer is: for p(t), as t gets larger the location of this subatomic particle will be some location further and further away (will get larger). This makes sense. Think of a car moving away from your house at a fixed speed traveling down a straight highway. At any given time, you could tell its location (let's say on a map) after so many minutes or hours. And, this location would be further and further away from your house as time elapsed.

Another solution to this equation is, once again starting with the original equation and simplify the negative exponent (-t) by taking the reciprocal of (-d^-t). This gives us (-1/d^t). So, we have a negative fraction (meaning either the numerator is negative or the denominator is negative). This time let's make the numerator negative which gives us: p(t) = (C^4-1) / (c^2 +1) +d^t. Now, as Sal shows us in the video, (c^4-1) = (c^2+1)(c^2-1) [the difference of 2 squares - this topic is taking up in algebra a little bit later). But, if you multiplied those 2 factors, you would end up with (c^4-1). Now, rewrite the equation as p(t) = (c^2+1)(c^2-1) / (c^2+1) + d^t. As, you see the (c^2+1) in the numerator cancels the (c^2+1) in the denominator leaving: p(t) = (c^2-1) / d^t. Once again, the numerator is a fixed constant while the denominator gets larger and larger as time (t) increases. This leaves the same result as in the first explanation (above): as the denominator increases the number represented by the fraction gets larger. • I am trying to listen to this and make sense of it. It has been 40 years since I studied mathematics in school and even though I have an MA in something else I can't make head of tails of this. Why is -d to the -t equal to 0? I don't remember ever learning about 0 exponents or negative exponents. I do remember that -d to the 2nd would be -d*-d. So I suppose that -d to the 1st would just be plain -d. But why does d to the 0 become 1 and not 0? • Can anyone help me with the next practice question, I'm flummoxed. The volume (in cubic meters), v, of a rectangular room is given by the expression:
v = 162-2b^6
Where b is a positive integer and each dimension is an integer greater than 1 meter. What are three unique expressions that could represent the dimensions of the room in terms of b?
In the answer box, write these three expressions as a product that equals the volume (e.g. "(expression 1)(expression 2)(expression 3)").

To be perfectly honest I don't even understand the question. But I came up with:
(27-b^2)(2+b)(2-b^2)
Thus my three unique expressions when all multiplied together add up to v but it's not quite there. Any advice or assistance would be great.  